Timeline for Surely recurrent random walks and the law of the iterated logarithm [closed]
Current License: CC BY-SA 3.0
20 events
| when toggle format | what | by | license | comment | |
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| Aug 25, 2016 at 22:28 | vote | accept | user45947 | ||
| Aug 25, 2016 at 17:53 | comment | added | Andrej Bauer | I missed the answer, sorry. I was going through the reviews, and I suspect it didn't show me the answer (or I just didn't scroll down). I am not an expert on this topic, but on the face of it the question looked all right to me. It is of course a judgment call whether a question is "not good" or just has a negative answer. | |
| Aug 25, 2016 at 17:25 | history | closed |
Anthony Quas Alexey Ustinov Franz Lemmermeyer Wolfgang Jan-Christoph Schlage-Puchta |
Not suitable for this site | |
| Aug 25, 2016 at 17:20 | comment | added | James Martin | (but also, parenthetically, if something is "not a good question" I had the impression that explaining the reason why in the comments, possibly along with a down-vote, is the recommended thing to do?) | |
| Aug 25, 2016 at 17:14 | comment | added | James Martin | @Andrej hasn't Nate Eldredge done that? | |
| Aug 25, 2016 at 16:38 | comment | added | Andrej Bauer | I propose that instead of having this question down-voted and possibly closed someone simply answer it negatively. @JamesMartin, perhaps? | |
| Aug 25, 2016 at 14:13 | answer | added | Nate Eldredge | timeline score: 3 | |
| Aug 25, 2016 at 14:00 | comment | added | James Martin | This is still hopeless. The mention of probability is a distraction; it plays no role in your question. If $f_n$ is any function that grows slower than linearly, you can find an up-down path $s_n$ that grows slower than linearly but satisfies $s_n=f_n$ for infinitely many values of $n$. To take an arbitrary example, $(+)^1(-)^3(+)^5(-)^7(+)^9....$ gives you $|s_{k^2}|=k$ for every $k$. | |
| Aug 25, 2016 at 12:50 | comment | added | user45947 | @AnthonyQuas I thought a bit more about the problem and added some specifications that should have been there at first hand. It might be that this does not change anything. Possibly a counterexample similar to yours can still be found. | |
| Aug 25, 2016 at 12:47 | history | edited | user45947 | CC BY-SA 3.0 |
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| Aug 25, 2016 at 11:58 | history | edited | user45947 | CC BY-SA 3.0 |
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| Aug 25, 2016 at 9:41 | review | Close votes | |||
| Aug 25, 2016 at 17:25 | |||||
| Aug 25, 2016 at 9:28 | comment | added | user45947 | Thanks. That's the kind of counterexample I wanted. Should probably have spent one more minute figuring that one out. | |
| Aug 25, 2016 at 9:26 | comment | added | Anthony Quas | Replace 0's by $-1$'s. | |
| Aug 25, 2016 at 9:24 | comment | added | user45947 | @AnthonyQuas Note that $X_i=\pm 1$. | |
| Aug 25, 2016 at 9:14 | comment | added | Anthony Quas | Still no. Consider (1,0,1,1,0,0,1,1,1,1,0,0,0,0,1,1,1,1,1,1,1,1,1,0,0,0,0,0,0,0,0,....) (that is $2^n$ 1's followed by $2^n$ 0's followed by $2^{n+1}$ 1's then $2^{n+1}$ 0's etc. For this one, the growth is linear in $n$ (not $\sqrt{n\log\log n}$). | |
| Aug 25, 2016 at 9:03 | comment | added | user45947 | I changed the equality to an inequality. That should account for all cases when $S_n$ is trivially bounded. I'm interested in whether there are any counterexamples where $S_n$ is surely recurrent, but infinitely many times has values larger than $\sqrt{2n \log \log n}$. This should be a null set, but it would be good to see a constructed example. | |
| Aug 25, 2016 at 8:57 | history | edited | user45947 | CC BY-SA 3.0 |
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| Aug 25, 2016 at 8:39 | comment | added | Nate Eldredge | Isn't the realization $(1,0,1,0,1,0,\dots)$ in your subset? I think if you write out your statements carefully, being explicit about the null sets, you'll see they aren't plausible. | |
| Aug 25, 2016 at 7:42 | history | asked | user45947 | CC BY-SA 3.0 |